Post on 16-Jul-2020
Course of Undergraduate Young Talent Project: 4-manifolds and Ricci FlowNotes, Part I. The proof of Hamilton’s 3-manifolds with positive Ricci curvature theorem
For lectures: Monday, March 26 - Wednesday, April 18, 2018
The basic Riemannian geometry section has been moved to the Appendix (last section).Hamilton’s papers are collected in the book [1].Hamilton’s papers will be cited as [3man] (1982), [4man] (1986).Perelman’s papers will be cited as [GP1], [GP2], and [GP3].
1 Variation formulas
Reference: §7 of [3man].
Lemma 1 If g (s) is a 1-parameter family of metrics with ∂∂sgij = vij (v is a symmetric covariant
2-tensor), then∂
∂sΓkij =
1
2gkℓ (∇ivjℓ +∇jviℓ −∇ℓvij) . (1)
Proof. The derivation of this formula illustrates a nice trick in computing evolutions of variousgeometric quantities such as the connection and the curvatures. We compute at an arbitrarily chosenpoint p ∈ M in normal coordinates {xi} centered at p so that Γk
ij (p) = 0. Note that ∂∂xi gjk (p) = 0.
In such coordinates, ∇kaj1···jqi1···ir (p) =
∂∂xka
j1···jqi1···ir (p) for any (r, q)-tensor a. Thus, at p we have
∂
∂sΓkij =
1
2gkℓ(
∂
∂xi
∂
∂sgjℓ +
∂
∂xj
∂
∂sgiℓ −
∂
∂xℓ
∂
∂sgij
)and hence (1) follows since
∇ivjℓ (p) =∂
∂xivjℓ (p)− Γm
ij (p) vmℓ − Γmiℓ (p) vjm =
∂
∂xivjℓ (p) .
Finally we note that since both sides of (1) are the components of tensors, equation (1) in fact holdsas a tensor equation, that is, it is true for any coordinate system, not just normal coordinates. Hereused the hint that the difference of two connections is a tensor and hence the variation of a family ofconnections is a tensor.
In the following one can skip the ∂∂s
Rm calculations if one just wants to prove Hamilton’s 3-manifold theorem because we will compute the evolution of the Einstein 2-tensor, which gives thesectional curvatures in dimension 3.
Lemma 2 If ∂∂sgij = vij, then
∂
∂sRℓ
ijk =1
2gℓp
{∇i∇jvkp +∇i∇kvjp −∇i∇pvjk
−∇j∇ivkp −∇j∇kvip +∇j∇pvik
}(2)
=1
2gℓp (∇i∇kvjp −∇i∇pvjk −∇j∇kvip +∇j∇pvik) (3)
− 1
2gℓp (Rijkqvqp +Rijpqvkq) .
1
Proof. FromRℓ
ijk = ∂iΓℓjk − ∂jΓ
ℓik + Γp
jkΓℓip − Γp
ikΓℓjp
we compute
∂
∂sRℓ
ijk = ∇i
(∂
∂sΓℓjk
)−∇j
(∂
∂sΓℓik
)=
1
2∇i
(∇jv
ℓk +∇kv
ℓj −∇ℓvjk
)− 1
2∇j
(∇iv
ℓk +∇kv
ℓi −∇ℓvik
),
which is the first formula. The second formula follows from this and the commutator formula
∇i∇jvℓk −∇j∇iv
ℓk = −Rq
ijkvℓq +Rℓ
ijqvqk.
Lemma 3 Suppose ∂∂sgij = vij. Then:
1.∂
∂sRij = −1
2
(∆Lvij +∇i∇jV −∇i (div v)j −∇j (div v)i
),
where the Lichnerowicz Laplacian is defined by
∆Lvij + ∆vij + 2Rkijℓvkℓ −Rikvjk −Rjkvik. (4)
2.∂
∂sR = ∇ℓ∇iviℓ −∆V − vij ·Rij.
Proof. (1) We have∂
∂sRij = ∇p
(∂
∂sΓpij
)−∇i
(∂
∂sΓppj
).
This follows from computing at the center in normal coordinates. Substituting the formula for ∂∂sΓkij
into this, we obtain∂
∂sRij =
1
2∇ℓ (∇ivjℓ +∇jviℓ −∇ℓvij)−
1
2∇i∇jV. (5)
Finally, commute derivatives.(2) Taking the trace, we obtain
∂
∂sR = gij
(∂
∂sRij
)− ∂
∂sgij ·Rij (6)
= ∇ℓ∇iviℓ −∆V − vij ·Rij.
We say that a 1-parameter family of metrics g(t) is a solution to the Ricci flow if ∂∂tgij = −2Rij.
Lemma 4 Let g(t) be a solution to the Ricci flow. Then:
2
1.∂R
∂t= ∆R + 2 |Rc|2 .
2.∂
∂tRij = ∆LRij = ∆Rij + 2RkijℓRkℓ − 2RikRjk.
3.
∂
∂tRijkℓ = ∆Rijkℓ + 2 (Bijkℓ −Bijℓk +Bikjℓ −Biℓjk) (7)
− (RipRpjkℓ +RjpRipkℓ +RkpRijpℓ +RℓpRijkp) ,
whereBijkℓ + −gprgqsRipjqRkrℓs = −RpijqRqℓkp. (8)
Proof. (1) and (2) follow immediately from the previous exercise.(3) We shall show the equivalent formula that under the Ricci flow, the (3, 1)-Riemann curvature
tensor evolves by
∂
∂tRℓ
ijk = ∆Rℓijk + gpq
(Rr
ijpRℓrqk − 2Rr
pikRℓjqr + 2Rℓ
pirRrjqk
)−Rp
iRℓpjk −Rp
jRℓipk −Rp
kRℓijp +Rℓ
pRpijk.
By applying the second Bianchi identity and commuting covariant derivatives, we compute
∆Rℓijk = gpq∇p∇qR
ℓijk = gpq∇p
(−∇iR
ℓjqk −∇jR
ℓqik
)= gpq
−∇i∇pR
ℓjqk +Rr
pijRℓrqk +Rr
piqRℓjrk +Rr
pikRℓjqr −Rℓ
pirRrjqk
+∇j∇pRℓqik −Rr
pjiRℓrqk −Rr
pjqRℓirk −Rr
pjkRℓiqr +Rℓ
pjrRriqk
and then apply the second Bianchi identity again to obtain
gpq∇pRℓjqk = gpqgℓm (−∇kRjqmp −∇mRjqpk) = ∇kR
ℓj −∇ℓRjk.
This lets us rewrite the formula for ∆Rℓijk given above as
∆Rℓijk = −∇i∇kR
ℓj +∇i∇ℓRjk +∇j∇kR
ℓi −∇j∇ℓRik
+ gpq
Rr
pijRℓrqk +Rr
piqRℓjrk +Rr
pikRℓjqr −Rℓ
pirRrjqk
−RrpjiR
ℓrqk −Rr
pjqRℓirk −Rr
pjkRℓiqr +Rℓ
pjrRriqk
.
Now the first Bianchi identity shows that
RrpijR
ℓrqk −Rr
pjiRℓrqk = −Rr
ijpRℓrqk,
which implies that ∆Rℓijk may be written as
∆Rℓijk = −∇i∇kR
ℓj +∇i∇ℓRjk +∇j∇kR
ℓi (9a)
−∇j∇ℓRik −RriR
ℓjrk +Rr
jRℓirk (9b)
+ gpq
−Rr
ijpRℓrqk +Rr
pikRℓjqr +Rℓ
pjrRriqk
−RℓpirR
rjqk −Rr
pjkRℓiqr
. (9c)
3
On the other hand, rewriting the commutator term gℓp (∇j∇i −∇i∇j)Rkp yields
∂
∂tRℓ
ijk = −∇i∇kRℓj +∇i∇ℓRjk +∇j∇kR
ℓi −∇j∇ℓRik (10a)
+ gℓp(Rq
ijkRqp +RqijpRkq
). (10b)
The reaction-diffusion equation for the Riemann curvature tensor now follows from comparing termsin formulas (9) and (10).
2 Uhlenbeck’s trick and the evolution of the Einstein tensor
Reference: §2 of [4man].The main goal is to prove (16). One can skip this section and simply first use (16) and prove it
later.Uhlenbeck’s trick (more concrete version for 3D; the n ≥ 4 case is essentially the same). Let
(M3, g(t)), t ∈ [0, T ), be a solution to the Ricci flow on a closed 3-manifold. By a standard result,M is parallelizable and so there exists a global orthonormal frame field {e0a}3a=1 for g(0) on M. Foreach x ∈ M let ea(x, t) be the solution to the ODE:
d
dtea(x, t) = Rcg(t)(ea(x, t)),
ea(x, 0) = e0a,
where we consider Rcg(t) : TxM → TxM as a (1, 1)-tensor. Then
d
dtg(t) (ea(t), eb(t)) = 0.
Hence {ea(t)}3a=1 is a global orthonormal frame field for g(t) on M for all t ∈ [0, T ).Let E = M× R3 be the trivial (flat) bundle with the Euclidean metric h on the fibers. Then we
have bundle isometriesι(t) : (E, h) → (TM, g(t))
defined by
ι(t) (x,v) =3∑
a=1
vaea(x, t),
where v = (v1, v2, v3). Note that ι(t)∗ (ea(x, t)) = ea, where ea is the standard Euclidean basis elementwith 1 in the a-th slot and 0 in the other slots. Of course, g(t)(ea(t), eb(t)) = h(ea, eb) = δab.
Exercise 5 For all dimensions, let ι0 : E → TM be a bundle isomorphism and let h = ι∗0(g(0)).Define bundle isomorphisms ι(t) : E → TM to solve the ODE:
d
dtι(t) = Rcg(t) ◦ι(t),
ι(0) = ι0.
Show that ι(t)∗(g(t)) is independent of t and hence is equal to h.
4
In the following, one can skip the evolution of Rabcd and instead just read the evolution of Eab.n ≥ 2. Now Rmg(t) is a section of the vector bundle Λ2T ∗M⊗S Λ
2T ∗M and the isomorphism ι(t)induces a pull back of this to Λ2E∗ ⊗S Λ2E∗, where the metric induced by g(t) is pulled back to themetric induced by h. Let R denote the pull-back of Rmg(t) by ι(t). We have
Rabcd = R(ea, eb, ec, ed)
= Rmg(t)(ea(t), eb(t), ec(t), ed(t)).
For the pulled back version of (7), we have the simplification:(∂
∂t−∆
)Rabcd = 2(Babcd − Babdc + Bacbd − Badbc), (11)
where Babcd = −RaebfRcedf .
Exercise 6 Prove (11).
With this, we can rewrite the evolution of R in a nice form. Define the square of R by
R2 = R ◦R : Λ2E∗ → Λ2E∗.
Define a Lie algebra structure on Λ2E∗ by
[U, V ]ab +∑c
(UacVcb − VacUcb)
for U, V ∈ Λ2E∗. Then Λ2E∗ ∼= so (n). Choose a basis {φα} of Λ2E∗ and let Cαβγ denote the structure
constants defined by[φα, φβ
]+∑
γ Cαβγ φγ. We define the Lie algebra square R# : Λ2E∗ → Λ2E∗ by
R#αβ = Cγδ
α Cεζβ RγεRδζ . (12)
Exercise 7 Show that∂
∂tR = ∆R+R2 +R#. (13)
When n = 3, we may express R2 +R# in the following simple way. Diagonalize R, i.e., choosean orthonormal basis {φα} of Λ2E∗ so that
(R(φα, φβ)
)=
λ 0 00 µ 00 0 ν
.
Then the matrix((R2 +R#)(φα, φβ)
)is also diagonal and
((R2 +R#)(φα, φβ)
)=
λ2 + µν 0 00 µ2 + λν 00 0 ν2 + λµ
. (14)
Here is an alternate (equivalent) way to obtain the 3D formula (14). With respect to an evolvingorthonormal frame {ea(t)} we have ∂
∂tgab = 0 and
∂
∂tRab = ∆Rab + 2RcabdRcd.
5
Hence Eab = Rgab − 2Rab (negative of the Einstein tensor) satisfies
∂
∂tEab = ∆Eab − 4RcabdRcd + 2 |Rc|2 gab.
Now assume n = 3. Then the eigenvalues of Eab and Rm are the same (twice the principal sectionalcurvatures). Moreover,
Rcabd = Rcdgab +Rabgcd −Rcbgad −Radgcb −1
2R (gcdgab − gcbgad)
Thus
∂
∂tEab = ∆Eab + 8RcbRac − 6RabR + 2
(R2 − |Rc|2
)gab
= ∆Eab +Qab,
where Qab = 2EcbEac − EabR +(R2 − 2 |Rc|2
)gab.
Let λ, µ, ν be the eigenvalues of Eab. Then λ+ µ+ ν = R. Since |Eab|2 = 4 |Rc|2 −R2 we have
R2 − 2 |Rc|2 = −1
2|Eab|2 +
1
2R2
= λµ+ λν + µν.
So
Q11 = 2E11E11 − E11R +(R2 − 2 |Rc|2
)g11
= 2λ2 − λ (λ+ µ+ ν) + (λµ+ λν + µν)
= λ2 + µν.
Similarly,
Q22 = µ2 + λν,
Q33 = ν2 + λµ.
This derives the evolution equation for Rm in dimension 3.For each x ∈ M, we may consider the ODE obtained from (13) by dropping the Laplacian term,
i.e., the ODE for R(t) : Λ2E∗x → Λ2E∗
x defined by
d
dtR = R2 +R#. (15)
Denoting the eigenvalues of R in the same way as for R, the ODE is (this also follows from theevolution of Eab):
dλ
dt= λ2 + µν,
dµ
dt= µ2 + λν,
dν
dt= ν2 + λµ.
Reference: §3 of [4man].The following is Hamilton’s maximum principle for systems. Let
V = Λ2E∗ ⊗S Λ2E∗.
6
Theorem 8 Let g (t), t ∈ [0, T ), be a solution to the Ricci flow on a closed manifold Mn. Let K ⊂ Vbe a subset which is invariant under parallel translation and whose intersection Kx + K ∩ Vx witheach fiber is closed and convex.Suppose the ode (15) has the property that for any R (0) ∈ K, we have R (t) ∈ K for all t ∈ [0, T ).If R (0) ∈ K, then R (t) ∈ K for all t ∈ [0, T ).
Let Rx ∈ Vx. For any path γ : [0, 1] → M with γ(0) = x, let R(u) be the parallel translation of Rx
along γ. Let vx ∈ Λ2E∗x be an eigenvector of Rx with eigenvalue λ. Let v(u) be the parallel translation
of vx along γ. Then R(u) (v(u)) is a parallel section of Λ2E∗ with R(0) (v(0)) = Rx(vx) = λvx. Bythe uniqueness of parallel translation, we have R(u) (v(u)) = λv(u) for u ∈ [0, 1]. So the eigenvaluesof an element R ∈ V are preserved under parallel translation. So if a subset K of V is defined byinequalities regarding eigenvalues of R, then K is invariant under parallel translation.
Note also that for any k ≥ 1, the minimum over all k-planes P in Λ2E∗ of trace(R|P ) is a concavefunction since for each P the function R 7→ trace(R|P ) is linear. The set of all R ∈ Vx where aconcave function of R is at least a constant c is a convex subset of Vx.
Sets defined by weak inequalities, i.e., ≥ or ≤, involving continuous functions of the eigenvaluesof R are closed.
For an exposition of the following, see chapter 3 of [4].
Exercise 9 Show that for any constant C0 ∈ R, if Rg(0) ≥ C0, then Rg(t) ≥ C0 for all t ∈ [0, T ). Thisis true for all n ≥ 3, but just prove this for n = 3. Hint: Let
K = {R ∈ V : λ(R) + µ(R) + ν(R) ≥ C0}
and show that ddt(λ+ µ+ ν) ≥ 0.
Answer : The set K is preserved by the ODE because
d
dt(λ+ µ+ ν) =
1
2
((λ+ µ)2 + (λ+ ν)2 + (µ+ ν)2
)≥ 0.
Exercise 10 n = 3. Show that if Rcg(0) ≥ 0, then Rcg(t) ≥ 0 for all t ∈ [0, T ). Hint: Order theeigenvalues by λ ≥ µ ≥ ν. Let
K = {R ∈ V : µ(R) + ν(R) ≥ 0} .
Note that µ+ ν is a concave function and
(µ+ ν) (R) = min{R(φ1, φ1
)+R
(φ2, φ2
):{φ1, φ2
}orthonormal
}.
This is the same as the minimum over all 2-planes P in Λ2E∗ of trace(R|P ).
Answer : The set K is preserved by the ODE because
d
dt(µ+ ν) = λ2 + µ2 + (µ+ ν)λ ≥ 0
whenever µ+ ν ≥ 0 (since λ ≥ µ ≥ 0).
Exercise 11 Suppose that M3 is closed and (M, g(0)) has positive Ricci curvature.
7
1. Show that there exists ε > 0 such that Rcg(0) ≥ εRg(0)g(0) on M. Show also that there existsC ≥ 1/2 such that λ
(Rmg(0)
)≤ C
(µ(Rmg(0)
)+ ν
(Rmg(0)
)).
2. Show that there exists ε > 0 such that Rcg(t) ≥ εRg(t)g(t) on M for all t ∈ [0, T ). Hint: GivenC ≥ 1/2, let
K = {R ∈ V : λ (R) ≤ C (µ (R) + ν (R))} ,where λ ≥ µ ≥ ν. Show that if λ−C (µ+ ν) = 0, then d
dt[λ− C (µ+ ν)] ≤ 0. (Note that λ (R)
for R ∈ Vx is the maximum of R(v, v) over all unit vectors v ∈ Λ2E∗x and hence is a convex
function of R.)
Answer : (1) is easy.(2)
d
dt(λ− C (µ+ ν)) = λ (λ− C (µ+ ν))− C
(µ2 − 1
Cµν + ν2
).
Hence, if λ− C (µ+ ν) = 0 and C ≥ 1/2 (i.e., 1C≤ 2), then
d
dt(λ− C (µ+ ν)) ≤ 0.
Exercise 12 Given C0 > 0, C1 ≥ 1/2, C2 < ∞ and δ > 0, let
K =
R ∈ V :λ(R)−ν(R)−C2 (λ(R)+µ(R)+ν(R))1−δ≤ 0,
λ (R) ≤ C1 (µ (R) + ν (R)) ,λ(R)+µ(R)+ν(R) ≥ C0
.
Show thatd
dtln
(λ− ν
(λ+ µ+ ν)1−δ
)≤ δ (λ+ ν − µ)− (1− δ)
µ2
λ+ µ+ ν.
Show that for δ > 0 small enough (e.g., δ1−δ
≤ 112C2
1works), we have
d
dtln
(λ− ν
(λ+ µ+ ν)1−δ
)≤ 0.
Conclude that there exist constants C < ∞ and δ > 0 such that∣∣∣∣Rc−1
3Rg
∣∣∣∣ ≤ CR1−δ. (16)
Note that (λ− ν) (R)− C ((λ+ µ+ ν) (R))1−δ is a convex function of R.
Answer : λ+ µ+ ν ≥ C0 and λ ≤ C1 (µ+ ν) are preserved. We compute
d
dtln
(λ− ν
(λ+ µ+ ν)1−δ
)
= δ (λ+ ν − µ)− (1− δ)(µ+ ν)µ+ (µ− ν)λ+ µ2
λ+ µ+ ν
≤ δ (λ+ ν − µ)− (1− δ)µ2
λ+ µ+ ν.
8
Note thatµ2
λ+ µ+ ν≥ 1
6
(µ+ ν)µ
λ≥ 1
6C1
µ
since λ+ µ ≤ 2λ ≤ 2C1 (µ+ ν) , and we also have
λ+ ν − µ ≤ λ ≤ C1 (µ+ ν) ≤ 2C1µ.
Hence, choosing δ > 0 small enough so that δ1−δ
≤ 112C2
1, we have
d
dtlog
(λ− ν
(λ+ µ+ ν)1−δ
)≤ 0.
Since λ− ν ≥∣∣Rc−1
3Rg∣∣ , we obtain (16).
3 Entropy monotonicity and no local collapsing
Reference: §3, §4, and §9 of [GP1]. (Exposition: chapter 6 of [3].)
∂
∂tgij = −2Rij, (17a)
∂f
∂t= −R−∆f + |∇f |2 + n
2τ, (17b)
dτ
dt= −1, (17c)
v =(τ(R + 2∆f − |∇f |2
)+ f − n
)τ−
n2 e−f .
W(g, f, τ) =
∫M
(τ(R + |∇f |2
)+ f − n
)τ−
n2 e−fdµ =
∫M
vdµ.
�∗ = − ∂∂t−∆+R.
−�∗(τ−n2 e−f ) =
(n
2τ− ∂f
∂t−∆f + |∇f |2 −R
)τ−
n2 e−f = 0.
−�∗v = 2τ
∣∣∣∣Rc+∇2f − 1
2τg
∣∣∣∣2 τ−n2 e−f . (18)
(18) implies
d
dtW(g(t), f(t), τ(t)) =
∫M
(∂
∂tv −Rv
)dµ
=
∫M
(−�∗v) dµ
= 2τ
∫M
∣∣∣∣Rc+∇2f − 1
2τg
∣∣∣∣2 τ−n2 e−fdµ.
9
Proof of (18). (∂
∂t+∆−R
)R = 2∆R + 2 |Rc|2 −R2
(∂
∂t+∆−R
)(∆f) = ∆
(∂f
∂t
)+ 2Rc ·∇2f + (∆−R) (∆f)
= −∆R +∆|∇f |2 + 2Rc ·∇2f −R∆f(∂
∂t+∆−R
)|∇f |2 = 2Rc(∇f,∇f) + 2∇f · ∇
(−R−∆f + |∇f |2
)+ (∆−R) |∇f |2
So
−�∗ (R + 2∆f − |∇f |2)= 2 |Rc|2 −R2 +∆|∇f |2 + 4Rc ·∇2f − 2R∆f
− 2Rc(∇f,∇f) + 2∇f · ∇(R +∆f − |∇f |2
)+R|∇f |2.
Using∆ |∇f |2 = 2 |∇∇f |2 + 2Rc(∇f,∇f) + 2∇f · ∇∆f,
this simplifies to
−�∗ (R + 2∆f − |∇f |2)= 2
∣∣Rc+∇2f∣∣2 + 2∇f · ∇
(R + 2∆f − |∇f |2
)−R
(R + 2∆f − |∇f |2
).
By −�∗(τ−n2 e−f ) = 0 and
−�∗(ab) = −b�∗a− a�∗b+ 2∇a · ∇b+Rab,
we have−�∗ ((R + 2∆f − |∇f |2
)τ−
n2 e−f
)= 2τ−
n2 e−f
∣∣Rc+∇2f∣∣2 .
By −�∗(τ−n2 e−f ) = 0 again and −�∗f = −Rf −R + |∇f |2 + n
2τ,
−�∗ (τ−n2 e−ff
)= τ−
n2 e−f (−�∗f) + 2∇
(τ−
n2 e−f
)· ∇f +Rτ−
n2 e−ff
= τ−n2 e−f
(−R− |∇f |2 + n
2τ
).
Finally,
−�∗ ((τ (R + 2∆f − |∇f |2)+ f − n
)τ−
n2 e−f
)= 2τ 1−
n2 e−f
∣∣Rc+∇2f∣∣2 − (R + 2∆f − |∇f |2
)τ−
n2 e−f
+ τ−n2 e−f
(−R− |∇f |2 + n
2τ
)= 2τ
∣∣∣∣Rc+∇2f − 1
2τg
∣∣∣∣2 τ−n2 e−f .
Logarithmic Sobolev inequality. Let (Mn, g) be a closed Riemannian manifold. Then for any a > 0there exists C(a, g) such that for any φ satisfying
∫φ2 = 1 we have∫
φ2 lnφ ≤ a
∫|∇φ|2 + C(a, g).
10
Entropy lower bound. Let w = τ−n4 e−
f2 , so that
∫w2 = 1 by assumption. Taking a = 2τ , we have
W (g, f, τ) =
∫M
(τ(Rw2 + 4 |∇w|2
)−(2 lnw +
n
2ln τ + n
)w2)dµ (19)
≥ τRmin −n
2ln τ − n− 2C(2τ, g)Vol(g).
µ(g, τ) = inf∫τ−
n2 e−f=1
W(g, f, τ),
ν(g) = infτ>0
µ(g, τ).
Scaling:
µ(g, τ) = µ (cg, cτ) ,
ν(g) = ν(cg).
By (19) we have µ(g, τ) > −∞.Standard elliptic theory implies the existence of a minimizer f of W(g, · , τ), and each such mini-
mizer satisfiesτ(R + 2∆f − |∇f |2
)+ f − n = µ (g, τ) .
Under the Ricci flow, if t1 ≤ t2, then
µ (g (t2) , τ (t2)) ≥ µ (g (t1) , τ (t1)) .
In particular,µ(g (t) , r2
)≥ µ
(g (0) , r2 + t
).
Nonrigorous proof of no local collapsing. Suppose R ≤ r−2 in B = Btr(x), where t ≥ T
2and r ≤ ρ.
Let f(t) = cχB. Then∫M r−ne−fdµ = 1 implies c = ln VolB
rn. By entropy monotonicity and since
T2< r2 + t ≤ ρ2 + T , we have (ignoring the |∇f |2 term)
lnκ + minτ∈[T
2,ρ2+T ]
µ (g (0) , τ)
≤ W(g (0) , f(0), r2 + t
)≤ W
(g (t) , f(t), r2
)≤ r2 max
BR + c− n
≤ lnVolB
rn.
See the aforementioned references for the actual proof.
4 Three-manifolds with positive curvature
Short time existence ([3man] (do not read), [DeTurck] (read)):
Theorem 13 If Mn is a closed Riemannian manifold and if g0 is a C∞ Riemannian metric, thenthere exists a unique smooth solution g (t) to the Ricci flow defined on some time interval [0, δ), δ > 0,with g (0) = g0.
11
The normalized Ricci flow is∂
∂tgij = −2Rij +
2
nrgij, (20)
where r = Vol (g)−1 ∫M Rdµ is the average scalar curvature.
Exercise 14 Show that ddtVol (g (t)) = 0 under the normalized Ricci flow.
Reference: [7], [2].
Theorem 15 If (M2, g0) is a closed Riemannian surface, there exists a unique solution g (t) of thenormalized Ricci flow (when n = 2, 2Rij = Rgij)
∂
∂tg = (r −R) g,
g (0) = g0.
The solution exists for all time. As t → ∞, the metrics g (t) converge uniformly in any Ck-norm toa smooth metric g∞ of constant curvature.
Reference: [3man].
Theorem 16 Let (M3, g0) be a closed Riemannian 3-manifold with positive Ricci curvature. Thenthere exists a unique solution g (t) of the normalized Ricci flow with g (0) = g0 for all t ≥ 0. Further-more, as t → ∞, the metrics g(t) converge exponentially fast in every Ck-norm to a C∞ metric g∞with constant positive sectional curvature. Hence M is diffeomorphic to a spherical space form.
Hamilton’s Cheeger–Gromov compactness theorem for solutions to the Ricci flow is discussed insection 5 below.
Hamilton’s point selection for obtaining singularity models. Reference: §16 of [8]Curvature blows up at least at the Type I rate. If the singular time T is finite, then
K (t) = maxx∈M
|Rm| (x, t) ≥ 1
8 (T − t)for all t ∈ [0, T ). (21)
Since ∂∂t|Rm|2 ≤ ∆ |Rm|2 + 16 |Rm|3, by the maximum principle, K (t) satisfies dK
dt≤ 16K
32 . The
result follows from limt→T K (t)−12 = 0.
Finite time singularity : Since ∂R∂t
= ∆R+ 2 |Rc|2 ≥ ∆R+ 23R2, Rmin(t) ≥
1
Rmin(0)−1 − 23t. Hence
a singularity forms at time T ≤ 32Rmin(0)
−1.Elementary point selection. We describe how to choose (xi, ti) so that the rescaled solutions
(M, gi (t) , xi), where
gi (t) = Kig(K−1
i t+ ti), Ki = |Rm| (xi, ti) , t ∈ [−Kiti, Ki (T − ti))
converge.
12
Type I. Let ti → T . By (21), there exist xi such that (T − ti)Ki ≥ 18. Since |Rm| (x, t) ≤ C
T−t, we
have
|Rm gi| (x, t) = K−1i |Rm g|
(x,K−1
i t+ ti)
≤ CK−1i
(T − ti)−K−1i t
≤ C18− t
for t ∈ [−Kiti,18). By this curvature bound, by no local collapsing, and by Hamilton’s Cheeger–
Gromov compactness theorem, (M, gi (t) , xi) converges to an ancient solution (M3∞, g∞(t), x∞), t ∈
(−∞, 18), satisfying |Rm g∞| (x, t) ≤ C
18−t.
Type II. Let Ti → T and choose (xi, ti) ∈ M× [0, Ti] so that
(Ti − ti)Ki = maxM×[0,Ti]
(Ti − t) |Rm| (x, t) ,
which → ∞ by the Type II assumption. Since |Rm| (x, t) ≤ (Ti−ti)Ki
Ti−t, we have
|Rm gi| (x, t) = K−1i |Rm g|
(x,K−1
i t+ ti)
≤ (Ti − ti)Ki
(Ti − ti)Ki − t
on M × [−tiKi, (Ti − ti)Ki). Since (Ti − ti)Ki → ∞, we have that (M, gi (t) , xi) converges to aneternal solution (M3
∞, g∞(t), x∞), t ∈ (−∞,∞), satisfying |Rm g∞ | (x, t) ≤ 1.In either the Type I or Type II case, the limit is a complete ancient solution with Rc g∞ ≥ 0
and |Rm g∞| (x∞, 0) = 1, in particular, Rg∞ (x∞, 0) > 0. Let U be the open subset of M∞ on whichRg∞ > 0. For the original solution g(t) there exists δ > 0 such that∣∣Rc−1
3Rg∣∣
R≤ δ−1R−δ. (22)
Since Ki → ∞, on U × (−∞, 0] we have|Rcg∞ − 1
3Rg∞g∞|
Rg∞≡ 0, which implies Rg∞ is constant by the
contracted second Bianchi identity. (For the rescaled sequence of solutions, (22) says|Rcgi −
13Rgigi|
Rgi≤
K−δi δ−1R−δ
gi, which limits to zero in U as i → ∞ since Rgi → Rg∞ > 0 in U .) This implies that
U = M∞. This implies that (M∞, g∞(0)) satisfies Rcg∞(0) =13Rg∞(0)g∞(0). Hence M∞ is compact,
which implies that it is diffeomorphic to M.
5 Cheeger–Gromov compactness theorem
Reference: [9]. (Exposition: chapters 3 and 4 of [3].)
Definition 17 Let K ⊂ M be an open set whose closure K is compact and let {gi} and g∞ beRiemannian metrics on K. Given k ≥ 0, we say that gi converges to g∞ in Ck on K if for everyε > 0 there exists i0 such that for i ≥ i0,
sup0≤α≤k
supx∈K
|∇α (gi − g∞)|g < ε,
13
where the covariant derivative ∇ is with respect to a fixed complete Riemannian metric g on K. Thisdefinition is independent of the choice of g.
Definition 18 Suppose {Ui} is an exhaustion of a smooth manifold Mn by open sets and gi areRiemannian metrics on Ui. We say that (Ui, gi) converges to (M, g∞) in C∞ on compact setsin M if for any subset K of M with compact closure and for each k > 0, {gi|K}i≥i0
converges to g∞|Kin Ck on K, where i0 is such that K ⊂ Ui for i ≥ i0.
C∞ pointed Cheeger–Gromov convergence:
Definition 19 A sequence {(Mni , gi (t) , xi)} , t ∈ (α, ω) , of complete pointed solutions to the Ricci
flow converges to a complete pointed solution to the Ricci flow (Mn∞, g∞ (t) , x∞) , t ∈ (α, ω) , if there
exist :
1. an exhaustion {Ui} of M∞ by open sets with x∞ ∈ Ui, and
2. a sequence of diffeomorphisms Φi : Ui → Vi = Φi (Ui) ⊂ Mi with Φi (x∞) = xi
such that(Ui × (α, ω) ,Φ∗
i
[gi (t)|Vi
]+ dt2
)converges to (M∞×(α, ω),g∞ (t)+dt2) in C∞ on compact
sets in M∞ × (α, ω) .
Hamilton’s Cheeger–Gromov compactness theorem for solutions to the Ricci flow:
Theorem 20 Let {(Mni , gi (t) , xi)} , t ∈ (α, ω) ∋ 0, be a sequence of complete pointed solutions to
the Ricci flow where there exists a constant C0 such that :
1.∣∣Rm gi(t)
∣∣ ≤ C0 on Mi × (α, ω),
2. Volgi(0)Bgi(0)
1 (xi) ≥ C−10 .
Then there exists a subsequence such that {(Mni , gi (t) , xi)} converges to a complete pointed solution
to the Ricci flow (Mn∞, g∞ (t) , x∞) , t ∈ (α, ω) , as i → ∞.
Example. Let (R× S1 (2) , h) denote the flat cylinder, where h = dx2 + dθ2 and θ ∈ S1 (2) =R/4πZ. The Rosenau solution is the solution g (t) = u (t)h to the Ricci flow defined for t < 0 by
u (x, t) =sinh (−t)
coshx+ cosh t. (23)
The metrics g (t) extend to smooth metrics on S2 by adding the north pole N and south pole S atthe two ends of R× S1 (2). From
Rg(t) = −∆h log u
u=
cosh t coshx+ 1
sinh (−t) (coshx+ cosh t),
one checks that ∂∂tgij = −Rgij = −2Rij.
Observe that for x ∈ R fixed,
limt→−∞
u (x, t) = limt→−∞
sinh (−t)
coshx+ cosh t= 1.
14
Hence, for any fixed p ∈ S2 − {N,S} (which corresponds to (x, θ) ∈ R × S1 (2)), we have that(S2, g(t), p) converges to (R× S1 (2) , h) as t → −∞.
On the other hand, we have
u (x+ t, t) = (− coshx coth t− sinhx− coth t)−1 ,
so thatlim
t→−∞u (x+ t, t) = (cosh x− sinhx+ 1)−1 =
(e−x + 1
)−1.
Let ϕt : R×S1 (2) → R×S1 (2) (ϕt extends to a map from S2 to itself which preserves both N and S)be defined by ϕt (x, θ) = (x+ t, θ) . Then (ϕ∗
tg) (x, t) = u (x+ t, t)h. Hence
limt→−∞
(ϕ∗t g) (x, t) =
(e−x + 1
)−1h (x) . (24)
Changing variables by x̃ = x/2 and θ̃ = θ/2, we have(e−x + 1
)−1h (x) = 4
(e−2x̃ + 1
)−1(dx̃2 + dθ̃2
), (25)
where x̃ ∈ R and θ̃ ∈ S1 (1) . We now show that this is 4 times the cigar metric. Recall
gcig =dx2 + dy2
1 + x2 + y2= ds2 + tanh2 s dθ̃2.
Let x̃ = ln(sinh s). Then dx̃ = coth s ds, so that ds2 = tanh2 s dx̃2. ex̃ = sinh s. So e−2x̃+1 = coth2 s.
Thus gcig =(e−2x̃ + 1
)−1(dx̃2 + dθ̃2
).
6 Compact singularity models
Reference: [13]. (Exposition: chapter 17, §1 of [3].)Energy. Define
λ(g) = inf∫e−fdµ=1
F(g, f),
where F(g, f) =∫ (
R + |∇f |2)e−fdµ. Scaling property: λ(cg) = c−1λ(g). It is easy to see
Rmin ≤ λ(g) ≤ Ravg,
where the upper bound follows from taking f = lnVol(g).Standard elliptic theory implies that there exists a C∞ minimizer f0 of F(g, f) under the constraint∫
e−fdµ = 1. Another standard argument implies that the minimizer is unique. The Euler–Lagrangeequation is:
2∆f0 − |∇f0|2 +R = λ(g).
Let g(t), t ∈ I (I is an interval), be a solution to the Ricci flow and let λ(t) = λ(g(t)). Given t0,let f0 be the minimizer for λ(t0). Solve ∂f
∂t= −R − ∆f + |∇f |2 for t ≤ t0 with f(t0) = f0. Since
15
λ(t) ≤ F(g(t), f(t)) + F(t) for t ≤ t0 and λ(t0) = F(t0), we have in the sense of the lim inf ofbackward difference quotients:
d
dt−λ(t)|t=t0 ≥
d
dtF(t)|t=t0
= 2
∫ ∣∣Rc+∇2f∣∣2 e−fdµ
≥ 2
n
∫(R +∆f)2 e−fdµ
≤ 2
n
(∫(R +∆f) e−fdµ
)2
=2
nF(t0)
2
=2
nλ(t0)
2
since∫e−fdµ = 1.
No breathers theorem: Suppose g(t) is such that there exist t1 < t2 and a diffeomorphism φ :M → M such that g(t2) = φ∗g(t1). g(t) is called a breather solution. Then λ(g(t2)) = λ(g(t1)). Letf2 be a minimizer for F(g(t2), · ). Solve ∂f
∂t= −R−∆f + |∇f |2 for t ∈ [t1, t2] with f(t2) = f2. Then
for t ∈ [t1, t2] we have∫e−f(t)dµg(t) = 1 and
λ(g(t2)) = F(g(t2), f2) ≥ F(g(t), f(t)) ≥ F(g(t1), f1) ≥ λ(g(t1)).
Since λ(g(t2)) = λ(g(t1)), we conclude that F(g(t), f(t)) = λ(g(t)) is constant for t ∈ [t1, t2]. Hencef(t) is a minimizer for F(g(t), · ) and d
dtF(g(t), f(t)) = 0, which implies Rcg(t) +∇2
g(t)f(t) = 0 for
t ∈ [t1, t2]. Thus g(t) is a steady GRS.
Entropy. Recall the entropy:
W(g, f, τ) =
∫ (τ(R + |∇f |2
)+ f − n
)τ−
n2 e−fdµ.
µ(g, τ) = inf∫τ−
n2 e−f=1
W(g, f, τ).
Let f̄ = f + n2ln τ , so that e−f̄ = τ−
n2 e−f and the constraint for W is equivalent to
∫e−f̄ = 1. Then,
under the constraint:
µ(g, τ) ≤ W(g, f, τ) = τ
∫(R + |∇f̄ |2)e−f̄dµ+
∫ (f̄ − n
2ln τ)e−f̄dµ− n. (26)
So taking f so that f̄ is the minimizer for λ(g), we obtain
µ(g, τ) ≤ τλ(g) + e−1Vol(g)− n
2ln τ − n
since xe−x ≤ e−1.Suppose λ(g) ≤ 0. Then
µ(g, τ) ≤ e−1Vol(g)− n
2ln τ − n.
16
In particular, limτ→∞ µ(g, τ) = −∞, so that ν(g) = −∞. Now, by scale-invariance, for any c > 0 wehave λ(cg) = c−1λ(g) ≤ 0 and
µ(g, τ) = µ(cg, cτ) ≤ e−1Vol(cg)− n
2ln(cτ)− n.
Taking c = Vol(g)−2n , we have Vol(cg) = 1, so that
µ(g, τ) ≤ ln(τ−
n2 Vol(g)
)+ e−1 − n ≤ lnVol(τ−1g).
Let g(t), t ∈ [0, T ), T < ∞, be a solution to the Ricci flow. Suppose that λ(g(t)) ≤ 0 for all t.Then taking τ = 1, we have
lnVol(g(t)) ≥ µ(g(t), 1) ≥ µ(g(0), 1 + t) ≥ −C = minτ∈[1,1+T ]
µ(g(0), τ) > −∞.
In particular, we cannot have limt→T Vol(g(t)) = 0.Suppose g(t), t ∈ [0, T ), T < ∞, is a singular solution to the Ricci flow with a compact singularity
model. Then limt→T Vol(g(t)) = 0, so that λ(g(t0)) > 0 for some t0. This also implies λ(g(t)) ≥λ(g(t0)) > 0 for t ≥ t0.
By (26), we have for any f satisfying the constraint that
W(g, f, τ) = τF(g, f̄) +
∫f̄ e−f̄dµ− n
2ln τ − n.
By the logarithmic Sobolev inequality (see Notes 3) with a = 2 and since∫e−f̄dµ = 1 (let φ = e−
f̄2 ),
we have ∫f̄ e−f̄dµ ≥ −
∫|∇f̄ |2e−f̄dµ− 2C(2, g)
≥ F(g, f̄) +Rmin − 2C(2, g).
Hence, for τ > 1 we have
µ(g, τ) = inffW(g, f, τ) ≥ (τ − 1)λ(g) +Rmin − 2C(2, g)− n
2ln τ − n.
Thus, if λ(g) > 0, then limτ→∞ µ(g, τ) = ∞.Claim. µ(g, τ) < 0 for τ sufficiently small.Proof of the claim. By the short time existence theorem, there exists δ > 0 and a unique solution
g(t), t ∈ [0, δ], to the Ricci flow with g(0) = g. Let u = τ−n2 e−f , where τ(t) = δ − t, be a solution
to �∗u = 0 with u = (4π)n2 δx0 for any x0 ∈ M, where δx0 is the Dirac delta function centered at
x0. Then limt→δ W(g(t), f(t), τ(t)) = 0. By entropy monotonicity, we have W(g(t), f(t), τ(t)) ≤ 0 fort ∈ [0, δ]. If W(g(0), f(0), τ(0)) = 0, then we are on a shrinking GRS with scale 0 at time δ, whereg(δ) is smooth. So Rm ≡ 0, which contradicts M being compact.
Claim. limτ→0 µ(g, τ) = 0. (We prove this later.)Since τ 7→ µ(g, τ) is continuous, we have
ν(g) = infτ>0
µ(g, τ) ∈ (−∞, 0).
Now since t 7→ ν(g(t)) is nondecreasing, we have limt→T ν(g(t)) + νT ∈ (−∞, 0] exists.
17
Theorem 21 (Zhenlei Zhang) Any compact singularity model is a shrinking GRS.
Proof. Suppose g(t), t ∈ [0, T ), T < ∞, is a singular solution to the Ricci flow with a compactsingularity model (Mn
∞, g∞(t)). Then there exist (xi, ti) with Ki = |Rm |(xi, ti) → ∞ such that(Mn
i , gi(t)) converges to (Mn∞, g∞(t)) and Mn
∞ is diffeomorphic to Mn. Hence λ(gi(t)) → λ(g∞(t)).Since λ(gi(t)) > 0, this implies λ(g∞(t)) ≥ 0.
Suppose there exists t0 such that λ(g∞(t)) = 0. Then λ(g∞(t)) = 0 for t ≤ t0. This implies g∞(t),t ≤ t0, is a steady GRS. Since Mn
∞ is compact, this implies Rcg∞(t) = 0. However, this contradictsno local collapsing: Let r be any positive real number. Since Rg∞(t) = 0 on Mn
∞, we have for any
x0 ∈ Mn∞ that Rg∞(t) ≤ r−2 in B = B
g∞(t)r (x0). Hence
Vol(Mn∞, g∞(t)) ≥ Vol(B, g∞(t)) ≥ κrn.
Since r is arbitrary, we have a contradiction.Finally, since after pullback gi(t) converges to g∞(t) pointwise in C∞, we have
ν(g∞(t)) = limi→∞
ν(gi(t)) = limi→∞
ν(g(K−1i t+ ti)) = νT
since for each t we have K−1i t + ti → T . Since ν(g∞(t)) is constant, we conclude that g∞(t) is a
shrinking GRS. This also implies that Mn admits a shrinking GRS structure. �
7 The proof of the strong maximum principle
Reference: §8 and §9 of [4man]. (Exposition: chapter 12 of [3].)We assume the following result:
Strong maximum principle (for the scalar heat equation). Let (Mn, g(t)), t ∈ I, be a family ofRiemannian metrics on a connected manifold, not necessarily complete. Suppose that u : M× I → Rsatisfies ∂u
∂t≥ ∆u and u ≥ 0. If u (x0, t0) = 0, then u(x, t) = 0 for all x and t ≤ t0.
Since ∂R∂t
≥ ∆R under the Ricci flow, if R ≥ 0 on M× I and R (x0, t0) = 0, then R(x, t) = 0 for
all x and t ≤ t0. Since∂R∂t
= ∆R + 2 |Rc|2, this implies Rc(x, t) = 0 for all x and t ≤ t0.
Strong maximum principle for Rm (statement). Let (Mn, g(t)), t ∈ [0, T ) be a solution to theRicci flow with nonnegative curvature operator on a connected manifold. Then for each t ∈ (0, T )the set image
(Rmg(t)
)⊂ Λ2T ∗M is a smooth subbundle which is invariant under parallel translation
(in space). Moreover, there exists a finite sequence of times 0 = t0 < t1 < t2 < · · · < tk < T suchthat image (Rm) (x, t) is a Lie subalgebra of Λ2T ∗
xM ∼= so (n) independent of time for t ∈ (ti−1, ti],1 ≤ i ≤ k, and for t ∈ (tk, T ). Furthermore, image(Rmg(0)) ⊂ image(Rmg(t1)) and image(Rmg(ti)) is aproper subalgebra of image(Rmg(tj)) for 1 ≤ i < j ≤ k and image(Rmg(tk)) is a proper subalgebra ofimage(Rmg(t)) for t > tk. If M is closed, then we may take k = 0.
Strong maximum principle for Rm (applications). We have the following immediate consequences:
1. If Rm = 0 at (x̄, t̄), then Rm (x, t) = 0 for all x ∈ M and t ≤ t̄.
2. If Rm > 0 at (x̄, 0), then Rm (x, t) > 0 for all x ∈ M and t > 0.
For n = 3 any subalgebra of so (3) besides 0 and so (3) is isomorphic to so (2).Let (M3, g(t)), t ∈ (α, ω), be a complete solution to the Ricci flow with bounded nonnegative
sectional curvature.
18
1. Suppose Rm = 0 at (x′, t′). Then the strong maximum principle implies Rm (x, t) = 0 for allx ∈ M and t ≤ t′. By the (forward) uniqueness theorem of Hamilton (compact case) and Chen andZhu (noncompact case), we have Rm = 0 on M× (α, ω).
2. Suppose image (Rm) (x′, t′) ∼= so (2). Then image (Rm) (x, t′) ∼= so (2) for x ∈ M is invari-ant under parallel translation. Lift the metric (M, g(t′)) to the universal cover (M̃, g̃(t′)). Then(M̃, g̃(t′)) = (N 2, h′) × R, where h′ has bounded curvature. By forward existence and uniqueness,we have for t ∈ [t′, ω) that (M̃, g̃(t)) is the product of R and some solution (N 2, h(t)). We cannothave Rm > 0 at any point at any time t < t′ since that would imply that Rm > 0 on M × [t′, ω),a contradiction. By (1), we also cannot have image (Rm) = 0 at any (x, t) ∈ M × (α, ω). Henceimage (Rm) (x, t) ∼= so (2) for all (x, t) ∈ M × (α, ω) is invariant under parallel translation andindependent of time. Hence (M̃, g̃(t)) is the product of R and a solution (N 2, h(t)) for t ∈ (α, ω).
3. Suppose Rm(x′, t′) > 0. Then Rm > 0 on M× (α, ω) for otherwise we would be in case (1) or(2), a contradiction.
To summarize, on all of M× (α, ω) either Rm = 0, Rm > 0, or the universal cover splits as theproduct of R and solution on a surface with positive curvature.
Strong maximum principle for Rm (proof in dimension 3). Let n = 3. Given a self-adjoint linearmap R : Λ2T ∗
xM → Λ2T ∗xM, let {λi (R)} denote the eigenvalues of R in nondecreasing order. Given
1 ≤ k ≤ 3, defineϕk (R) = λ1 (R) + · · ·+ λk (R) = inf
dimS=ktrace(R|S),
where the infimum is taken over all k-dimensional subspaces S of Λ2T ∗xM. In each fiber ϕk is a
concave function of R. Define ϕk : M× [0, T ) → R by
ϕk (x, t) = ϕk (Rm (x, t)) .
Lemma. If ϕk (x0, t0) > 0 for some k and x0 ∈ M, then ϕk > 0 on M× (t0, T ).
Proof of the lemma. Choose any smooth function h0 : M → R with 0 ≤ h0 ≤ ϕk and h0(x0) > 0.Let h : [t0, T ) → R be the solution to ∂h
∂t= ∆g(t)h with h(t0) = h0. By the strong maximum principle,
h > 0 on M× (t0, T ). So the lemma follows from:Claim. ϕk(x, t) ≥ h(x, t) on M× [t0, T ).Proof of the claim. Consider the system of PDE
∂
∂t(Rm, h) = ∆(Rm, h) + (Q, 0) .
Let K = {(R,h) |ϕk (R) ≥ h}. This a closed convex set which is invariant under parallel translation.The associated ODE is
d
dt(R,h) = (Q(R), 0),
where Q(R) = R2 +R#. Since Q(R) ≥ 0 when R ≥ 0, we have that K is preserved by the ODE.Hence K is preserved by the PDE. �
The null space of Rm at (x, t) is
null (Rm (x, t)) ={V ∈ Λ2T ∗
xM : Rm (x, t) (V,W ) = 0 for all W ∈ Λ2T ∗xM
}.
Since Rm ≥ 0,null (Rm (x, t)) =
{V ∈ Λ2T ∗
xM : Rm (x, t) (V, V ) = 0}.
19
By the lemma, there exist times 0 = t0 < t1 < t2 < · · · < tk < tk+1 = T such that dimnull (Rm (x, t))is constant on M× (ti−1, ti], 1 ≤ i ≤ k, and on M× (tk, T ).
Now suppose Rm11 (x0, t0) = 0, i.e., λ = 0. Then Q11 = 0, i.e., λ2 + µν = 0, so that µν = 0. ThusRm (x0, t0) = 0 or image(Rm (x0, t0)) is 1-dimensional.
If ϕ3 (x0, t0) = 0 somewhere, then ϕ3 = 0 on M× (α, t0]. By this and forward uniqueness, we haveRm = 0 on M× (α, ω).
So we may assume ϕ3 > 0 on M×(α, ω). For each (x0, t0) we have ϕ1 (x0, t0) > 0 or ϕ1 (x0, t0) > 0,which is equivalent to ϕ2 (x0, t0) = 0.
If ϕ1 > 0 on M× (α, ω), then Rm > 0 on M× (α, ω).Otherwise ϕ2 (x0, t0) = 0 somewhere, which implies ϕ2 = 0 on M×(α, t0]. This implies that either
ϕ2 = 0 on M× (α, ω) or there exists t1 ∈ (α, ω) such that ϕ2 = 0 on M× (α, t1] and ϕ1 > 0 (Rm > 0)on M× (t1, ω).
Claim. null(Rm) is invariant under parallel translation a subset of null(Q).
Proof of the claim. Let V (x, t) be a smooth local section of the subbundle null (Rm (x, t)) definedon an open subset of M× (α, ω). Using ∆ (Rm (V, V )) = 0 and Rm (∆V, V ) = 0, we compute that
0 =∂ Rm
∂t(V, V ) + 2Rm
(∂V
∂t, V
)(27)
= (∆Rm+Q) (V, V )
= −2gij Rm(∇iV,∇jV )− 4gij (∇i Rm) (∇jV, V ) +Q (V, V ) .
Since0 = ∇ (Rm (V )) = (∇Rm) (V ) + Rm (∇V ) ,
we have0 = gij (∇iRm) (∇jV, V ) + gij Rm(∇iV,∇jV ) .
Hence (27) yields2gij Rm(∇iV,∇jV ) +Q (V, V ) = 0.
Since both terms in the above equation are nonnegative, we conclude thatQ (V, V ) = 0 and gij Rm(∇iV,∇jV ) =0. In particular, null (Rm) ⊂ null (Q) and Rm (∇Y V,∇Y V ) = 0 for all Y , that is,
∇Y V (x, t) ∈ null (Rm (x, t)) for all Y. (28)
Now prove (28) implies that null (Rm (x, t)) is invariant under parallel translation. Let p =dimnull (Rm (x, t)), which is independent of x. Given (x0, t0), let {Wi(x)}pi=1 be a basis of smoothsections of null (Rm (x, t0)) defined in a neighborhood U of x0 in M and let γ : [0, L] → U be anypath with γ(0) = x0. Since ∇γ′(s)Wi ∈ null (Rm (γ(s), t0)), there exist functions Aj
i (s) such that
∇γ′(s)Wi =
p∑j=1
Aji (s)Wj.
Let V0 ∈ null (Rm (x0, t0)). Then V0 =∑p
i=1 fi0Wi (x0). Let V (s) =
∑pi=1 f
i(s)Wi(γ(s)). The IVP:
0 = ∇γ′(f iWi
)=
p∑i=1
(df i
dsWi + f i∇γ′Wi
), f i (0) = f i
0,
20
is equivalent to IVP for the system of linear ODE:
df i
ds+ f j
p∑j=1
Aij(s) = 0, f i (0) = f i
0,
which has a unique solution. Hence the parallel translate V (s) of V0 along γ is in null (Rm (γ(s), t0)).Finally we prove that null (Rm (x, t)) is constant in time onM×(ti−1, ti), 1 ≤ i ≤ k+1. Let V (x, t)
be a smooth local section of null (Rm (x, t)) defined in a space-time neighborhood in M× (ti−1, ti).Claim. V (x, t) ∈ null (∆Rm(x, t)).Proof of the claim. Since Rm (∇jV ) = 0, we have
0 = gij∇i (Rm (∇jV )) (29)
= gij (∇i Rm) (∇jV ) + Rm (∆V ) .
Since ∇eiV ∈ null(Rm) for any vector field ei, we have ∇ei (∇eiV ) ∈ null(Rm) and ∇∇eieiV ∈
null(Rm). Thus ∆V (x, t) ∈ null (Rm (x, t)), which implies
gij (∇i Rm) (∇jV ) = 0.
Using the above, we compute
0 = ∆ (Rm (V ))
= (∆Rm) (V ) + Rm (∆V ) + 2gij (∇iRm) (∇jV )
= (∆Rm) (V ) ,
which proves the claim.Now, since V ∈ null (Q (Rm)), we have
0 =∂
∂t(Rm (V ))
= (∆Rm+Q (Rm)) (V ) + Rm
(∂V
∂t
)= Rm
(∂V
∂t
).
Hence ∂V∂t
∈ null (Rm (t)). This implies null (Rm (t)) is constant in time.
8 Appendix: Review of Riemannian geometry.
Let (Mn, g) be a Riemannian metric and let ∇ denote the Levi-Civita connection. Given localcoordinates {xi}ni=1, the Christoffel symbols are the components of ∇: ∇ ∂
∂xi
∂∂xj = Γk
ij∂
∂xk .
Lemma 22
Γkij =
1
2gkℓ(
∂
∂xigjℓ +
∂
∂xjgiℓ −
∂
∂xℓgij
). (30)
21
Proof. For any vector fields X, Y and Z that
2g (∇XY, Z) = X (g (Y, Z)) + Y (g (X,Z))− Z (g (X, Y )) (31)
+ g ([X,Y ] , Z)− g ([X,Z] , Y )− g ([Y, Z] , X) .
Applying this to X = ∂∂xi , Y = ∂
∂xj , Z = ∂∂xℓ , multiplying bygkℓ, and using
[∂∂xi ,
∂∂xj
]= 0, we obtain
equation (30).Let α be a covariant r-tensor and X be a vector field. Lie derivative of α with respect to X is
denoted by LXα. Formula:
(LXα) (Y1, . . . , Yr) = (∇Xα) (Y1, . . . , Yr) (32)
+n∑
i=1
α (Y1, . . . , Yi−1,∇YiX,Yi+1, . . . , Yr) .
Since ∇Xg = 0, the Lie derivative of the metric is given by
(LXg) (Y1, Y2) = g (∇Y1X, Y2) + g (Y1,∇Y2X) . (33)
Lemma 23 If α is a covariant 2-tensor, then
(LXα)ij = ∇Xαij + gkℓ (∇iXkαℓj +∇jXkαiℓ)
and in particular(LXg)ij = ∇iXj +∇jXi,
so that if f is a function, then (Lgradg fg
)ij= 2∇i∇jf. (34)
Proof. Apply (32) with Y1 = ∂∂xi , Y2 = ∂
∂xj . To obtain equation (34), let Xi = ∇if . Note that∇i∇jf = ∇2f
(∂∂xi ,
∂∂xj
)and ∇2f (X, Y ) = X (Y f)− (∇XY ) f .
Let Rijkℓ = gℓmRmijk be the components of the Riemann curvature 4-tensor. The first and second
Bianchi identities are
Rijkℓ +Rjkiℓ +Rkijℓ = 0, (35)
∇iRjkℓm +∇jRkiℓm +∇kRijℓm = 0. (36)
Lemma 24 Let Rij be the Ricci tensor and let R be the scalar curvature. We have the contractedsecond Bianchi identity :
2gij∇iRjk = ∇kR. (37)
If g is an Einstein metric, i.e., Rij =1nRgij, and n ≥ 3, then R is a constant.
Proof. Multiplying (36) by gim, i.e., tracing, we obtain
0 = gim (∇iRjkℓm +∇jRkiℓm +∇kRijℓm)
= gim∇iRjkℓm −∇jRkℓ +∇kRjℓ.
Tracing again, i.e., multiplying by gkℓ, we obtain
gim∇iRjm −∇jR + gkℓ∇kRjℓ = 0,
22
which yields (37).Taking the divergence of Rjk =
1nRgjk yields
1
2∇kR = gij∇iRjk = gij∇i
(1
nRgjk
)=
1
n∇kR
and the result follows from n ̸= 2.The Ricci identities:
(∇i∇j −∇j∇i)αk1···kr = −r∑
ℓ=1
Rmijkℓ
αk1···kℓ−1mkℓ+1···kr . (38)
As special cases of (38), we have:
Lemma 25 1. If α is a 1-form, then
(∇i∇j −∇j∇i)αk = −Rℓijkαℓ.
2. If β is a 2-tensor, then
∇i∇jβkℓ −∇j∇iβkℓ = −Rpijk βpℓ −Rp
ijℓ βkp. (39)
Acting on functions, the Laplacian is
∆f = div∇f = trace g(∇2f) = gij∇i∇jf = gij(
∂2f
∂xi∂xj− Γk
ij
∂f
∂xk
).
One can show that
∆f =1√|g|
n∑i,j=1
∂
∂xi
(√|g|gij ∂f
∂xj
), (40)
where |g| + det (gij).
Lemma 26 1.∆∇if = ∇i∆f +Rij∇jf. (41)
2.∆ |∇f |2 = 2 |∇∇f |2 + 2Rij∇if∇jf + 2∇if∇i (∆f) . (42)
3. If Rc ≥ 0, ∆f ≡ 0, and |∇f | = 1, then ∇2f = ∇(∇f) = 0, i.e., ∇f is parallel, andRc (∇f,∇f) = 0.
Proof. (1) This follows from
∆∇if = ∇j∇i∇jf = ∇i∇j∇jf −Rjijk∇kf,
where we used the Ricci identities.(2) We compute
∆ |∇f |2 = ∇i∇i |∇jf |2 = 2∇i (∇i∇jf∇jf)
= 2 |∇i∇jf |2 + 2∆∇jf∇jf
23
and obtain part (2) from part (1).(3) If Rc ≥ 0 and ∆f = 0, then part (2) implies
∆ |∇f |2 = 2∣∣∇2f
∣∣2 + 2Rc (∇f,∇f) ≥ 2 |∇∇f |2 .
Hence, if also |∇f | = 0, then |∇2f | = 0 and Rc (∇f,∇f) = 0.Divergence theorem:
Lemma 27 Let (Mn, g) be a compact oriented Riemannian manifold. Let dµ denote the volume formof (M, g), let dσ denote the area form of ∂M, and let ν denote the unit outward normal to ∂M. IfX is a vector field, then ∫
Mdiv (X) dµ =
∫∂M
⟨X, ν⟩ dσ, (43)
where div (X) = ∇iXi. In particular, if M is closed, then
∫M div (X) dµ = 0.
Proof. Define the (n− 1)-form α byα + ιX(dµ).
Using d2 = 0 and LX(dµ) = (d ◦ ιX + ιX ◦ d) (dµ), we compute
dα = d ◦ ιX(dµ) = (d ◦ ιX + ιX ◦ d) (dµ) = LX (dµ) = div(X)dµ,
where to obtain the last equality, we may compute in an orthonormal frame e1, . . . , en :
LX (dµ) (e1, . . . , en) =n∑
i=1
dµ (e1, . . . ,∇eiX, . . . , en)
= div(X)dµ (e1, . . . , en) .
Now Stokes’s theorem implies that∫M
div(X)dµ =
∫M
dα =
∫∂M
α =
∫∂M
ιX(dµ) =
∫∂M
⟨X, ν⟩ dσ.
To verify the last equality, we used
(ιX(dµ)) (e2, . . . , en) = n (dµ) (X, e2, . . . , en)
= n ⟨X, ν⟩ (dµ) (e1, e2, . . . , en)
=1
(n− 1)!⟨X, ν⟩ .
Lemma 28 1. On a closed manifold,∫M ∆udµ = 0.
2. On a compact manifold,∫M
(u∆v − v∆u) dµ =
∫∂M
(u∂v
∂ν− v
∂u
∂ν
)dσ.
In particular, on a closed manifold∫M
u∆vdµ =
∫M
v∆udµ.
24
3. If f is a function and X is a 1-form, then∫M
f div (X) dµ = −∫M
⟨∇f,X⟩ dµ+
∫∂M
f ⟨X, ν⟩ dσ.
Proof. (1) This follows from the divergence theorem and ∆u = div(∇u).(2) This follows from the divergence theorem applied to X = u∇v − v∇u which has div(X) =
u∆v − v∆u.(3) This follows from the divergence theorem applied to the vector field fX.An easy calculation using coordinates where gij = δij at a point shows that for any 2-tensor aij,
|aij|2g ≥1
n
(gijaij
)2.
More generally, for a p-tensor a, p ≥ 2,∣∣ak1···kp∣∣2 ≥ 1
n
∣∣gijaijk3···kp∣∣2 .Let Rc denote the Ricci tensor.
Lemma 29 Show that on a closed manifold∫M
∣∣∇2f∣∣2 dµ+
∫Mn
Rc (∇f,∇f) dµ =
∫M
(∆f)2 dµ. (44)
Since |∇2f |2 ≥ 1n(∆f)2 , this implies∫
MRc (∇f,∇f) dµ ≤ n− 1
n
∫M
(∆f)2 dµ. (45)
Proof. This follows from (42) since integration by parts yields∫M ∇f ·∇∆fdµ = −
∫M (∆f)2 dµ.
Lichnerowicz inequality:
Lemma 30 Suppose f is an eigenfunction of the Laplacian with eigenvalue λ > 0:
∆f + λf = 0.
If Rc ≥ (n− 1)K, where K > 0 is a constant, then
λ ≥ nK. (46)
Proof. If ∆u+ λu = 0, then
1
2∆ |∇u|2 = |∇∇u|2 + ⟨∇∆u,∇u⟩+Rc (∇u,∇u)
≥ 1
n(∆u)2 − λ |∇u|2 + (n− 1)K |∇u|2
=1
nλ2u2 + ((n− 1)K − λ) |∇u|2 .
Integrating this, we have
0 ≥ 1
nλ2
∫M
u2dµ+ ((n− 1)K − λ)
∫M
|∇u|2 dµ.
Dividing by∫Mu2dµ and using the fact that λ =
∫M|∇u|2 dµ
/∫Mu2dµ implies
0 ≥ 1
nλ2 + ((n− 1)K − λ)λ.
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